Not clear about the change of basis in new space

Actually after I wrote down the query on the invertible matrix which I posted a few days ago I happened to refer again to Kunze Huffman and found that this is a standard theorem regarding transformation of linear operator from one basis to another.

Then I realized that the point which was not clear was that if Tv is a vector in basis B then how could with respect to B' I could write P(Tv) where P is the matrix of transformation from B to B'. What is unclear is that when u are doing this u are actually trying to premultiply a vector which is already in the space W, but according to theorem 8 on pg 53 of the book it says that

Suppose I is an n x n invertible matrix over F. Let V be an n dimensional vector space over F and let B be an ordered basis of V . Then there exists unique ordered basis B' of V such that [alpha] in basis B= P[alpha ] in basis B'.

So how is it here the vector space in which the vector is going to reside and the basis are completely different. Am I missing something very obvious. My thinking is that probably even if the space W has smaller dimension than V it is extended by adding 0s to it to equate V and then trying to apply the above technique. Still I am highly confused of applying a matrix NxN which would transform V -> V on something in W.

Sorry but I do not know how to use the subscripts here for clarity but hopefully I have been able to make my doubt across.

So how is it here the vector space in which the vector is going to reside and the basis are completely different.

It's possible for a vector space to have an infinite number of basis. For example, take the vector space [itex] \mathbb{R}^2[/itex]. Now [itex]B_1=\{ (0,1) , (1,0) \} [/itex] is a basis for the vector space. Notice that [itex]B_2=\{ (1,2) , (2,1) \} [/itex] is also a basis for the same vector space, and so on.

Then I realized that the point which was not clear was that if Tv is a vector in basis B then how could with respect to B' I could write P(Tv) where P is the matrix of transformation from B to B'.

When we represent a linear transform by a matrix, remember that it's only meaningful relative to some ordered basis. So, you can represent the same linear transform with respect to a different ordered basis.

It's possible for a vector space to have an infinite number of basis. For example, take the vector space [itex] \mathbb{R}^2[/itex]. Now [itex]B_1=\{ (0,1) , (1,0) \} [/itex] is a basis for the vector space. Notice that [itex]B_2=\{ (1,2) , (2,1) \} [/itex] is also a basis for the same vector space, and so on.

When we represent a linear transform by a matrix, remember that it's only meaningful relative to some ordered basis. So, you can represent the same linear transform with respect to a different ordered basis.

Did that help in clarifying your doubt?

The points which u mention is clear but what is unclear is the following What is unclear is that when u are doing this u are actually trying to premultiply a vector which is already in the space W. So does it mean if W ism dim space and P is n x n and m < n then when u multiply the P with Tv do u assume that the u extend the dimension of a vector in W to n by adding n -m 0 s to the end